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Theory For Undergraduates Problems Problem 1Boundary conditions for open strings.

Consider two static D2-branes in four dimensional spacetime (ct, x 1 , x 2 , x 3 ). The first one is at x 3 = 0. The second one is parallel to the first and is located at x 3 = a > 0. Sketch the branes. Consider open strings with σ ∈ [0, σ1 ] that stretch from the second brane (σ = 0) to the first brane (σ = σ 1 ).

State the boundary conditions (Free or Dirichlet, with value) for the string coordinates X µ ( t , σ∗) (list the eight conditions – µ = 0, 1, 2, 3 and σ∗ = 0, σ 1 ).

Problem 2. Spaces constructed by identifications.

Give a (simple!) fundamental domain F and describe the resulting space M plane z = x + iy:

for each of the following (single) identifications acting on the complex

(a) z ∼ z + i. (b) z ∼ 2z.

Problem 3Variation of an action Consider the Chern-Simons action for three-dimensional electromagnetism:

S =

∫

dt

∫

.

Σ

d 2 x A 0 F 1 2 + A 1 F 2 0 + A2F01 .

Recall that the field strength F µ ν = ∂ µ A ν − ∂ν A µ . Find the equation of motion resulting from the variation of the gauge field component A 0 (as usual, ignore boundary terms). The equation of motion can be written fully in terms of field strengths.

Problem 4.How heavy is a cosmic string?

A nearby relativistic cosmic string of tension T0 produces a cylindrical gravitational lens in which two images of a single faraway source would be separate by an angle δ = 8πGT0 .

(1)

This formula is given in units where c and � are set equal to one, the angle δ is measured in radians, and G c 6.7 × 10−11 m3/kg · s2 is Newton’s constant (c = 3 × 108m/s, � = 1.06 × 10−34 kg.m2/s.) (c) Complete (1) by adding whatever factors of c and/or � are needed. (d) A string produces the plausible value of δ = 0.5 arc-seconds (degree = 60 arc-minutes, arc- minute = 60 arc-seconds). What is the linear mass liveexamhelper.com density of such string in kg/m?

Problem 5. Angular momentum of a rotating open string.

An open string of length A and energy E rotates rigidly with angular velocity ω. Recall that ω A = c and A = 2 E .π T0 (a) Introduce a radial length r along the string and let d r denote a small piece of string a distance r from the center. What is the magnitude dp of the (relativistic) momentum carried by this small piece of string ? What is the magnitude d J of the angular momentum carried by this small piece of string? Both answers should be in terms of ω, T 0 , r, d r and constants. (b) Use integration to calculate the total angular momentum carried by the rotating string. Give your answer in terms of the energy E of the string and the string tension T0. ∫ 1 √x 2dx π .4 0 1−x 2 = Useful integral: Problem 6.Momentum of closed strings. For a free closed string we have X˙ ( t , σ ) =

.1 Σ F˙ ( u ) + ˙G(v ) , with u = ct + σ , v = ct − σ .

2

(1)

(a) Demonstrate that the periodicity condition σ ∼ σ + σ 1 (σ 1 = E/T 0 ) relates the lack of periodicity of F˙(u) to the lack of periodicity of G˙(v). (b) We now write F˙(u) = f ˙ (u)+ α˙ u ,

and G˙(v) = g˙(v)+ β˙ v ,

˙ 0 where f˙ and ˙g are strictly periodic functions with period σ 1 and α˙ and β˙ are constant vectors. does result in (a)total relate α˙ and ˙τ How T ∂X . the Calculate the 2∂t c (c) The momentum density (per unit σ) carried of the string is P = β ˙?

Plug back in (1) to find X˙(t, σ) in terms of f˙(u), ˙g(v), α˙, ct, and possibly σ. momentum p˙ carried by the string in terms of the vector α˙ and other constants.

Problem 7. You learned that a closed string stretched along a circle and released with zero initial velocity will contract to zero size at some later time. Consider a closed string that is stretched along an ellipse and is released with zero initial velocity. Will it contract to zero size? If yes, why? if not, why not? A complete answer requires a precise justification but, in fact, no calculation.

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Solutions Question 1

z = x³ = 0

y

 = ₁

= x² τ , 0), xX =1 (x¹ 0

X ( τ , 0), X 2 ( τ , 0),

X 0(τ, σ ) X 1 ( τ , σ 11) ⎬ X 2 ( τ , σ1) ⎭

X 3 ( τ , 0) = a X 3 ( τ , σ1 ) = 0

⎫ all “Free”

�

Dirichlet

Question 2 y i x

(a) z ∼z + i (b)

F :

0≤ y < 1

M :

infinite cylinder

z ∼ 2z

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This scales z, keeping the argument unchanged.

F : 1 ≤ |z| < 2 M :

torus

Technically we should also include the point z = 0.

1

2

Question 3 S =

�

δS =

�

�

=

d 3 x (A 0 F 12 + A 1 F 2 0 + A 2 F 0 1 )

d 3 x (δA 0 F 12 + A 1 (∂ 2 δA 0 ) − A2 ∂ 1 (δA0 )) F 12 = 0 d3 x (δA 0 F 12 − δA 0 ∂2 A 1 + (∂ 1 A2 )δA 0 )

Question 4 δ = 8πGT0 �

δS have = d 3 units xδA 0 with (2F 1 2suitable ) EOM:c’s and �’s δ should no (a) Units of GT 0?

F =

GM

2

r2

[G] = [F ] L 2= M M

[G] = [T 0] =

2

L L2 T2M 2

T L23M ML T2

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→ [GT0] = 4 T δ = 8πGT0

c2

T0

c2

4

thus need a c

c4

(b) Mass density µ T0

L4

= δ·

T 02 c

=

0

=

c2

8πG = 1.30 × 10

� 0.5

20

!

� �2 � � 3 × 10 8 �2π 360 · 60 · 60 �2 � π · 4 · (6 .67 × 10

− )11

kg m

Question 5

dr

r (a) v = ωr dp =

(d m )ωr ω2 r 2

� but dp =

−

c2

1 dE T dr dm = = 0 2 c2 c

T 0 ω�

c2 d J = rdp

rdr 1

=

− T0 ω

c2

(b)

�

1

ω2 r 2 c2

r2 dr −

ω2 r 2 c2

� l/ 2

T 0ω r2 dr � ω2 r 2 0 c2 1 − c2 c ωr let: r = ωx c = x � 3 x2 dx 1� J = 2 T0ω · c � √ 0 1 − x2 ω c2 J =2

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T0 c ω2

J =2

�1 0

�

√

x2 dx

=

π T0c

2 ω2 1 −x π� � � 4 l 2c ω = c ω = l 2 2c πT 0 c ω = 2E = E T 02 π T 0πcE 2 π 2T2 02c

J =

Question 6

= 2

2

2

E 0c πT

� 1� � F ( � u) + G(�v) 2

X� ( t , σ) = (a)

X� (t, σ + σ 1 ) = X� (t, σ) �(u, (u + −G F� (v (u)−= σG1 � G�+ (v −(v) σ1) FF� +σσ11)) + ) =(v) F�−(u) G�

(b) f�(u + σ 1 ) + α� (u + σ 1 ) − f�(u) − α� (u) = �g(v) + β�v − � 1� X�( t , σ) = f ( � u) + α� u + u) + α� 2 � v � �g( α� σ 1 = βσ�=1 1α� (u +→v) +α� 1f= (β� u) + �g(v) 2 2 � 1 �� X�( t , σ) = α� ct + f ( u) + �g(v) 2

�

�

�g(v − σ 1 ) + β�(v − σ1)

(c) T 0∂�x c 2 ∂t � T0 T0 �� P�τ = α� + f (ct + σ) + g ( �� ct − σ) 2c � �σ1 c �σ1 T0 ∂ T0 � � P�τ dσ = σ P c1α� + � 2c 0 0 ∂σ f�(ct + σ) − �g(ct − σ) dσ � 0, by periodicity� of �f and g � P�τ=

P� = Since σ 1 =

E

T0σ1

c α�

, Twe 0 can also write: E

P� = c α�

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Question 7

E₂ O E₁

P

Consider an arbitrary point P on the string. If each crawler travels L/4, the two endpoints E 1 and E 2 are separated along the ellipse by L/2. Moreover, they are opposite, and the midpoint of 4c �L 1 to the origin after time . �4 c E 1 E 2 is the origin. Thus P will go to the origin at t = L 1 . Since P is arbitrary, the ellipse collapses Comment: You may convince yourself that any curve C with inversion symmetry in the plane (if �a ∈ C, then (−�a) ∈ C) will collapse to zero size.

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